# How do you graph and solve |4x – 3|+ 2 <11?

Jul 16, 2018

The solution is $x \in \left(- \frac{3}{2} , 3\right)$

#### Explanation:

The inequality is

$| 4 x - 3 | + 2 < 11$

$| 4 x - 3 | - 9 < 0$

The point to consider is

$4 x - 3 = 0$

$\implies$, $x = \frac{3}{4}$

There are $2$ intervals to consider

$\left(- \infty , \frac{3}{4}\right)$ and $\left(\frac{3}{4} , + \infty\right)$

Therefore,

In the first interval

$- 4 x + 3 - 9 < 0$

$\implies$, $- 4 x - 6 < 0$

$\implies 0$, $4 x > - 6$

$\implies$, $x > - \frac{3}{2}$

This solution belongs to the interval

In the second interval

$4 x - 3 - 9 < 0$

$\implies$, $4 x - 12 < 0$

$\implies$, $x < 3$

This solution belongs to the interval

The solution is $x \in \left(- \frac{3}{2} , 3\right)$

graph{|4x-3|-9 [-20.27, 20.27, -10.14, 10.14]}